Question

The coefficient of $x^k$ in the expansion of $\frac{1-2x-x^2}{e^{-x}}$ is

• A. $\frac{1-k-k^2}{k!}$
• B. $\frac{k^2+1}{k!}$
• C. $\frac{1-k}{k!}$
• D. $\frac{1}{k!}$

Answer 1

Answer: (A)

Now, $\frac{1-2x-x^2}{e^{-x}}$
$=\left(1-2x-x^2\right)\left(e^x\right)$
$=\left(1-2x-x^2\right)(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots$
$+\frac{x^k}{k!}+\dots \infty )$
$=\left(1+x+\frac{x^2}{2!}+\dots +\frac{x^k}{k!}+..\infty \right)-2(x+x^2+\frac{x^3}{2!}$
$+\dots +\frac{x^k}{(k-1)!}+\frac{x^{k+1}}{k!}+\dots \infty )$
$-\frac{x^k}{k!}+\dots \infty )$
$-(x^2+x^3+\frac{x^4}{2!}+\dots .$
$\frac{x^k}{(k-2)!}+\frac{x^{k+1}}{(k-1)!}+\frac{x^{k+2}}{k!}+\dots \infty )$
$\therefore$ Coefficient of $x^k$ in $\left(\frac{1-2x-x^2}{e^{-x}}\right)$
$=\frac{1}{k!}-\frac{2}{(k-1)!}-\frac{1}{(k-2)!}$
$=\frac{1}{k!}-\frac{2k}{k(k-1)!}-\frac{k(k-1)}{k(k-1)(k-2)!}$
$=\frac{1}{k!}-\frac{2k}{k!}-\frac{k(k-1)}{k!}$
$=\frac{1-k-k^2}{k!}$